The interval in which the function f(x) = x^x | vedclass

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(D) Given the function $f(x) = x^x$ for $x > 0$.Taking the natural logarithm on both sides,we get $\ln(f(x)) = x \ln(x)$.Differentiating both sides wi

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$\left[\frac{1}{e}, \infty\right)$

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(D) Given the function $f(x) = x^x$ for $x > 0$.Taking the natural logarithm on both sides,we get $\ln(f(x)) = x \ln(x)$.Differentiating both sides with respect to $x$ using the product rule:$\frac{1}{f(x)} f'(x) = 1 \cdot \ln(x) + x \cdot \frac{1}{x} = \ln(x) + 1$.Thus,$f'(x) = x^x(1 + \ln(x))$.For the function to be strictly increasing,we require $f'(x) > 0$.Since $x^x > 0$ for all $x > 0$,the condition $f'(x) > 0$ implies $1 + \ln(x) > 0$.$\ln(x) > -1$.$x > e^{-1}$,which means $x > \frac{1}{e}$.Therefore,the interval in which the function is strictly increasing is $\left(\frac{1}{e}, \infty\right)$.Note: The option $\left[\frac{1}{e}, \infty\right)$ is the standard representation for the interval where the function is non-decreasing.

The interval in which the function $f(x) = x^x, x > 0$,is strictly increasing is

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